Monday Math 109

Last time, we found the formula for the Laplace transform of the derivative of a function; if we have f(t), t>0, with Laplace transform F(s), then.

Now, let us consider what happens to Laplace transforms as s→∞. By the definition of the Laplace transform, . Noting that as for all t>0, we see that in the limit as s→∞, the integrand goes to zero throughout the entire region of integration, except the point t=0; so long as f(t) does not have a delta function spike at t=0, the integral has to go to zero in the limit. Thus,.
So, let us take the limit as s→∞ of our derivative formula. So long as f(t) has no step discontinuities at t=0, f’(t) will not have a delta function spike at t=0, and the above applies to its transform, so we see that.
This is called the initial value theorem (since it involves the initial value of f).
To find the version for when there is a discontinuity at t=0, we have to remove the point at zero from the integral (replace lower limit 0 with 0+, and so we get.

Next, let us take the limit as s→0 of the Laplace transform of f’(t). When s=0, , and so,
and so, if we take the limit as s→0 of our derivative formula,,
this last being the final value theorem. This tells us that the behavior of f(t) as t→∞ is reflected in the location of the poles of F(s); specifically, we see four cases:
I. If all poles are on the left side () of the s complex plane, then F(0) exists and is finite, and so ; our function f(t) decays to zero exponentially (or faster) at large t.
II. If there are poles are on the right side () of the s complex plane, such as for the Laplace transforms of the hyperbolic sine and cosine, then f(t) contains terms growing exponentially, and f(∞) does not exist.
III. If there are complex conjugate poles on the imaginary axis () of the s complex plane, such as for the Laplace transforms of the sine and cosine, then f(t) contains sinusoidal terms, and f(∞) is undefined.
IV. If there is a simple pole at the origin of the s complex plane (and none in the right side), such as for the Laplace transform of the constant function, then exists and is nonzero.

3 Responses to “Monday Math 109”

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